Q.A ball is dropped from a height of $90\ \text{m}$ on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between $t = 0$ to $12\ \text{s}$.

The key idea is Projectile Motion with Inelastic Collisions.

We use the acceleration due to gravity $g = 10\ \text{m/s}^2$.

1. First Fall: The ball falls from $H_0 = 90\ \text{m}$.

   • Speed just before first collision: $v_1 = \sqrt{2gH_0} = \sqrt{2 \times 10 \times 90} = \sqrt{1800} = 30\sqrt{2}\ \text{m/s}$.
   • Time for first fall: $t_1 = v_1/g = (30\sqrt{2})/10 = 3\sqrt{2}\ \text{s} \approx 4.24\ \text{s}$.

2. First Bounce Cycle:

   • Speed after first collision: $v_1' = v_1 - (1/10)v_1 = (9/10)v_1 = (9/10)(30\sqrt{2}) = 27\sqrt{2}\ \text{m/s}$.
   • Time for this upward and downward journey: $T_1 = 2v_1'/g = 2(27\sqrt{2})/10 = 5.4\sqrt{2}\ \text{s} \approx 7.64\ \text{s}$.
   • Time of second collision: $t_1 + T_1 = 3\sqrt{2} + 5.4\sqrt{2} = 8.4\sqrt{2}\ \text{s} \approx 11.88\ \text{s}$.

3. Second Bounce and Final Segment:

   • Speed after second collision: $v_2' = (9/10)v_1' = (9/10)(27\sqrt{2}) = 24.3\sqrt{2}\ \text{m/s}$.
   • The total time $11.88\ \text{s}$ is just before $12\ \text{s}$. For the remaining $12 - 11.88 = 0.12\ \text{s}$, the ball moves upwards. …

The problem asks us to plot the speed-time graph of a ball dropped from a height, where it loses a fraction of its speed at each collision with the floor. This scenario involves understanding kinematics under constant gravitational acceleration and the effect of inelastic collisions.

Concept and Intuition

The motion of the ball is a sequence of vertical free falls and ascents.

1. Free Fall (Downward Motion): When the ball is falling, its speed increases linearly due to gravity. If it starts from rest, its speed $v$ at time $t$ is given by $v = gt$. The displacement $s$ is $s = \frac{1}{2}gt^2$.
2. Vertical Ascent (Upward Motion): After bouncing, the ball moves upwards. Its speed decreases linearly due to gravity, eventually becoming zero at its maximum height. If it starts with an upward speed $v_0$, its speed $v$ at time $t$ is $v = v_0 - gt$. The time to reach maximum height is $t_{max} = v_0/g$, and the maximum height reached is $H_{max} = v_0^2/(2g)$.
3. Inelastic Collision with Floor: This is the critical aspect. The problem states that the ball loses one-tenth of its speed at each collision. This means if the speed just before collision is $v_{\text{before}}$, the speed just after collision $v_{\text{after}}$ will be $v_{\text{after}} = v_{\text{before}} - \frac{1}{10}v_{\text{before}} = \frac{9}{10}v_{\text{before}}$. This loss of speed signifies a loss of kinetic energy, making the collision inelastic.
4. Consequences of Speed Loss: Because the ball rebounds with less speed, it will not reach its original height. The maximum height achieved after each subsequent bounce will be lower. Consequently, the time taken for each subsequent ascent and descent cycle will also be shorter.
5. Speed-Time Graph Characteristics:
   • Since speed is the magnitude of velocity, the graph will always be above or on the time axis (speed is non-negative).
   • During free fall, speed increases linearly with time, so the graph segment will have a positive slope ($+g$).
   • During vertical ascent, speed decreases linearly with time, so the graph segment will have a negative slope ($-g$).
   • At the moment of collision, there is an instantaneous drop in speed, represented by a vertical line segment on the graph.
   • The overall graph will be a series of "V" shapes, with each subsequent "V" being shorter and narrower due to the energy loss.

We will use the standard acceleration due to gravity $g = 10\ \text{m/s}^2$ for calculations, as is common in many Indian competitive exams unless specified otherwise. We will also use $\sqrt{2} \approx 1.414$ for numerical approximations.

Step-by-Step Solution

We need to track the ball's speed and time through each phase of its motion until $t = 12\ \text{s}$.

1. First Fall (from $t=0$ until the first collision)

The ball is dropped from a height of $H_0 = 90\ \text{m}$.

• Time to hit the floor ($t_1$): Using the equation $H = ut + \frac{1}{2}gt^2$, with $u=0$:

Numerically, $t_1 \approx 3 \times 1.414 = 4.242\ \text{s}$.

• Speed just before the first collision ($v_1$): Using $v = u + gt$:

Numerically, $v_1 \approx 30 \times 1.414 = 42.42\ \text{m/s}$.

This segment of the graph goes from $(0, 0)$ to $(4.242\ \text{s}, 42.42\ \text{m/s})$.

2. First Rebound and Ascent (after the first collision)

• Speed just after the first collision ($v_1'$): The ball loses one-tenth of its speed.

Numerically, $v_1' \approx 27 \times 1.414 = 38.178\ \text{m/s}$.
At $t_1 \approx 4.242\ \text{s}$, the speed instantaneously drops from $42.42\ \text{m/s}$ to $38.178\ \text{m/s}$.

• Time to reach maximum height ($t_{\text{asc1}}$): The ball moves upwards with initial speed $v_1'$. At maximum height, its speed is $0$. Using $v = u - gt$:

Numerically, $t_{\text{asc1}} \approx 2.7 \times 1.414 = 3.8178\ \text{s}$.

• Total time at the peak of the first bounce ($T_1$):

Numerically, $T_1 \approx 5.7 \times 1.414 = 8.0598\ \text{s}$.

This segment of the graph goes from $(4.242\ \text{s}, 38.178\ \text{m/s})$ to $(8.0598\ \text{s}, 0\ \text{m/s})$.

3. Second Fall (from $T_1$ until the second collision)

The ball falls from rest at height $H_1$ (reached after the first bounce). The time taken to fall from $H_1$ is equal to the time taken to ascend to $H_1$.

• Time for the second fall ($t_{\text{fall2}}$):

Numerically, $t_{\text{fall2}} \approx 3.8178\ \text{s}$.

• Speed just before the second collision ($v_2$): This speed will be equal to the speed with which it started its first ascent ($v_1'$).

Numerically, $v_2 \approx 38.178\ \text{m/s}$.

• Total time at the second collision ($T_2$):

Numerically, $T_2 \approx 8.4 \times 1.41421 = 11.88\ \text{s}$.

This segment of the graph goes from $(8.06\ \text{s}, 0\ \text{m/s})$ to $(11.88\ \text{s}, 38.18\ \text{m/s})$.

4. Second Rebound and Ascent (partial, up to $t=12\ \text{s}$)

The second collision occurs at $T_2 \approx 11.88\ \text{s}$. We need to plot the graph up to $t=12\ \text{s}$.

• Speed just after the second collision ($v_2'$):

Numerically, $v_2' \approx 24.3 \times 1.41421 = 34.37\ \text{m/s}$.
At $T_2 \approx 11.88\ \text{s}$, the speed instantaneously drops from $38.18\ \text{m/s}$ to $34.37\ \text{m/s}$.

• Time remaining for ascent:

Numerically, $\Delta t \approx 12 - 11.8794 = 0.1206\ \text{s}$.

• Speed at $t=12\ \text{s}$ ($v(12)$): The ball is ascending with initial speed $v_2'$. …

Concept: A Closed-Form (Geometric Series) Formula for Repeated Inelastic Bounces

Method: General Recursive/Closed-Form Formula, Evaluated at $n=1,2$ — Not Step-by-Step Arithmetic

Each bounce multiplies the ball's speed by a fixed ratio $r = 9/10$ (loses one-tenth of its speed). Because the time between successive bounces also scales by this same ratio, the cumulative time up to the $n$-th bounce is a geometric series with a known closed form — letting you jump straight to "which bounce interval contains $t=12\ \text{s}$" instead of computing each bounce's numbers in sequence.

Steps

1. Time and speed of the very first impact. Dropped from $H=90\ \text{m}$, $g=10\ \text{m s}^{-2}$:

2. Set up the recursion. After the $k$-th bounce, rebound speed is $r^k v_1$ (where $v_1$ is the impact speed at the first bounce, and $r=0.9$). The time for the "up-then-down" cycle following the $k$-th bounce is $2\times(\text{rebound speed})/g = \dfrac{2r^k v_1}{g} = 2 r^k t_1$.

3. Sum the geometric series to get the time of the $n$-th bounce, $T_n$:

This single formula gives the time of any bounce directly — no need to add up each preceding cycle one at a time.

4. Evaluate for $n=2$ (the second bounce), with $r=0.9$, $t_1\approx4.2426\ \text{s}$:

5. Evaluate for $n=3$ to bracket $t=12\ \text{s}$:

Since $T_2\approx11.88 < 12 < T_3\approx18.75$, the instant $t=12\ \text{s}$ falls in the rising (post-2nd-bounce) phase. …